How do you find the #lim_(xrarr3) (x^2+x-12)/(x-3)#?

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Answer 1 · 2017-04-16T01:34:48

#7#

Rewrite the numerator as #(x+4)(x-3)#

#lim_(x->3) ((x+4)(x-3))/(x-3)#

Cancel the #x-3#

#lim_(x->3) x+4#

Plug in #3# for #x#:

#3+4=7#

Answer 2 · 2017-04-16T01:47:38

The numerator factors and cancels the denominator, leaving a simple linear expression that can be evaluated at the limit.

Given: #lim_(xrarr3) (x^2+x-12)/(x-3)#

Factor the numerator:

#lim_(xrarr3) ((x-3)(x+4))/(x-3)#

Please observe the factors that cancel:

#lim_(xrarr3) (cancel(x-3)(x+4))/cancel(x-3)#

This leaves a simple linear expression that can be evaluated at the limit:

#lim_(xrarr3) x+4 = 7#

Answer 3 · 2017-04-16T18:47:55

#7#

#"factorise numerator and simplify"#

#lim_(xto3)((x+4)(cancel(x-3)))/(cancel(x-3))#

#=lim_(xto3)(x+4)=3+4=7#